9 Perturbation Theory II: Time Dependent Case

In this chapter we’ll study perturbation theory in the case that the perturbation varies

in time. Unlike the time–independent case, where we mostly wanted to know the bound

state energy levels of the system including the perturbation, in the time–dependent case

our interest is usually in the rate at which a quantum system changes its state in response

to the presence of the perturbation.

9.1 The Interaction Picture

Consider the Hamiltonian

H(t) = H0 + ∆S(t) (9.1)

where H0 is a time-independent Hamiltonian whose spectrum and eigenstates we under-

stand, and ∆S(t) is a perturbation that we now allow to depend on time. (The footnote

on ∆S(t) is to remind us that this operator has genuine, explicit time dependence even in

the Schrodinger picture.) The idea is that H0 models a system in quiescent state, and we

want to understand how eigenstates of H0 respond as they start to notice the perturbation.

In the absence of the time-dependent perturbation, the time evolution operator would be

U0(t) = e−iH0t/~ as appropriate for the model Hamiltonian H0. We define the interaction

picture state |ψI(t)〉 as

|ψI(t)〉 = U−10 (t)|ψS(t)〉 , (9.2)

where |ψS(t)〉 is the Schrodinger picture state evolving in accordance with the TDSE for

the full Hamiltonian (9.1). Thus, in the absence of ∆S(t), we’d simply have |ψI(t)〉 =

U−10 (t)U0(t)|ψ(0)〉 = |ψ(0)〉 so the state |ψI(0)〉 would in fact be independent of time. The

presence of the perturbation means that our states do not evolve purely according to U0(t),

so |ψI(t)〉 has non-trivial time dependence.

To calculate the evolution of |ψI(t)〉, we differentiate

i~∂

∂t|ψI(t)〉 = −H0 eiH0t/~|ψS(t)〉+ eiH0t/~ i~

∂

∂t|ψS(t)〉

= −H0 eiH0t/~|ψS(t)〉+ eiH0t/~ (H0 + ∆(t))|ψS(t)〉= U−1

0 (t) ∆(t)U0(t)|ψI(t)〉 ,

(9.3)

where we’ve used the full TDSE to evaluate ∂|ψS〉/∂t, and in the last line we’ve written

our state back in terms of the interaction picture. Similarly, the expectation value of an

operator AS in the Schrodinger picture becomes

〈ψS(t)|AS|ψS(t)〉 = 〈ψI(t)|U−10 (t)ASU0(t)|ψI(t)〉 (9.4)

in terms of the interaction picture states, so we define the interaction picture operator

AI(t) by

AI(t) = U−10 (t)ASU0(t) (9.5)

again using just the evolution operators of the unperturbed, rather than full, Hamiltonian.

Infinitesimally, this is

d

dtAI(t) =

i

~[H0, AI(t)] + U−1

0 (t)∂AI(t)

∂tU0(t) (9.6)

– 128 –

where the final term comes from the explicit time dependence present even in the Schrodinger

picture operator. In this sense, the interaction picture is ‘halfway between’ the Schrodinger

picture, in which states evolve and operators are time independent, and the Heisenberg pic-

ture, in which states are always taken at their initial value |ψ(0)〉, but operators evolve

according to the full Hamiltonian. In the interaction picture, states evolve by just the

time-dependent perturbation as

i~∂

∂t|ψI(t)〉 = ∆I(t)|ψI(t)〉 (9.7)

where ∆I(t) = U−10 (t)∆(t)U0(t) is the perturbation in the interaction picture, whilst oper-

ators also evolve, but only via the model Hamiltonian.

It will turn out to be useful to introduce an interaction picture time evolution operator

UI(t) defined so that

|ψI(t)〉 = UI(t)|ψI(0)〉 for any initial state |ψI(0)〉 ∈ H . (9.8)

From (9.3) we find that UI(t) satisfies the differential equation

i~∂

∂tUI(t) = ∆I(t)UI(t) . (9.9)

Note that since the operator ∆I(t) itself depends on time, we cannot immediately integrate

this to write UI(t) in terms of the exponential of ∆I(t): in particular it is generally not

correct to claim

UI(t)?= exp

(− i

~

∫ t

0∆I(t

′) dt′), (9.10)

because since the operator ∆I(t) itself now depends on time, in general

[∆I(t),∆I(t′)] 6= 0 .

Thus, when expanding the exponential as a power series, we’d have to commute the oper-

ators at different times through one another.

So far our considerations have been exact, but to make progress we must now approx-

imate. To do so, it’ll be convenient to rewrite (9.9) as an integral equation77

UI(t) = 1− i

~

∫ t

0∆I(t

′)UI(t′) dt′ (9.11)

in which the operator UI(t) we wish to understand also appears on the rhs inside the

integral. We now replace the UI(t′) in the integral (9.11) by its value according to its own

equation, obtaining

UI(t) = 1− i

~

∫ t

0∆I(t1) dt1 +

(− i

~

)2 ∫ t

0∆I(t1)

[∫ t1

0∆I(t2)UI(t2) dt2

]dt1 . (9.12)

77To obtain this equation, we’ve just integrated (9.9) using the initial condition UI(0) = 1.

– 129 –

The term containing UI(t) on the rhs now has two explicit powers of the perturbation

∆(t), so may be hoped to be of lower order in our perturbative treatment. Iterating this

procedure gives

UI(t) =∞∑n=0

(− i

~

)n ∫ t

0dt1 · · ·

∫ tn−1

0dtn ∆I(t1) ∆I(t2) · · ·∆I(tn) (9.13)

in general. Note that the operators ∆(ti) at different times do not necessarily commute —

they appear in this expression in a time–ordered sequence, with the integrals taken over

the range

0 ≤ tn ≤ tn−1 ≤ · · · ≤ t1 ≤ t .Instead of the naıve expression (9.10), we often write the series (9.13) as

UI(t) = T exp

(− i

~

∫ t

0∆I(t

′) dt′)

(9.14)

for short, where the time–ordered exponential 78 T exp(· · · ) is defined by the series (9.13).

It’s a good exercise to check that this time–ordered exponential is equal to the usual

exponential in the case that [∆(t1),∆(t2)] = 0 for all t1 and t2.

The interaction picture perturbation is ∆I(t) = U−10 (t)∆S(t)U0(t), so we can also

write (9.13) as

UI(t) =∞∑n=0

(− i

~

)n t∫0

dt1 · · ·tn−1∫0

dtn U−10 (t1)∆S(t1)U0(t1−t2) ∆S(t2) · · ·U0(tn−1−tn) ∆S(tn)U0(tn)

(9.15)

using the fact that U−10 (ti)U0(ti−1) = U0(ti−1− ti) . In this form we see that the perturba-

tive expansion of UI(t) treats the effect of the perturbation as a sequence of impulses: we

evolve according to H0 for some time tn ≥ 0 then feel the effect ∆S(tn) of the perturbation

just at this time, before evolving according to H0 for a further time (tn−1 − tn) ≥ 0 and

feeling the next impulse from the perturbation, and so on. Finally, we integrate over all

the possible intermediate times at which the effects of the perturbation were felt, and sum

over how many times it acted. This method, closely related to Green’s functions in pdes,

is also the basis of Feynman diagrams in QFT, where H0 is usually taken to be a free

Hamiltonian.

Now let’s put these ideas to work. Suppose that at some initial time t0 we’re in the

state |ψ(t0)〉 =∑

n an(t0)|n〉, described in terms of some superposition of the basis |n〉of eigenstates of the unperturbed Hamiltonian. Then a later time t the interaction picture

state will be |ψI(t)〉 = UI(t − t0)|ψ(t0)〉 and can again be described by a superposition∑n an(t)|n〉. Contracting with a state 〈k| gives

ak(t) = 〈k|UI(t− t0)|ψI(t0)〉 (9.16)

78Time–ordered exponentials were introduced by Freeman Dyson and play a prominent role in QFT.

They’re also familiar in differential geometry in the context of computing the holonomy of a connection

around a closed path. (This is not a coincidence, but it would take us too far afield to explain here.)

– 130 –

in the interaction picture. Thus, using (9.13) we have

ak(t) ≈ ak(t0)− i

~∑n

an(t0)

∫ t

t0

〈k|∆I(t′) |n〉 dt′

= ak(t0)− i

~∑n

an(t0)

∫ t

t0

〈k|U−10 (t′)∆S(t′)U0(t′) |n〉 dt′

= ak(t0)− i

~∑n

an(t0)

∫ t

t0

ei(Ek−En)t′/~ 〈k|∆S(t′)|n〉 dt′ ,

(9.17)

to first non-trivial order in ∆, where we used the fact that H0|n〉 = En|n〉. In particular,

if at t = t0 we begin in an eigenstate |m〉 of H0, so that

an(t0) =

1 if n = m

0 else, (9.18)

then the amplitude for our perturbation to caused the system to make a transition into a

different H0 eigenstate |k〉 6= |m〉 is

ak(t) = − i

~

∫ t

t0

ei(Ek−Em)t′/~ 〈k|∆S(t′)|m〉 dt′ (9.19)

to first order in the perturbation. Henceforth, we’ll drop the subscript on the Schrodinger

picture perturbation – note that this is the form in which the perturbation enters the

original Hamiltonian.

To go further, we must specify how the perturbation ∆(t) actually depends on time.

9.2 Prodding the Harmonic Oscillator

As a simple example, let’s consider a d = 1 harmonic oscillator that receives a gentle

‘prod’ described by the time-dependent force Fpr(t) = F0 e−t2/τ2 , in addition to the force

F = −kX from the spring constant. We can describe the kick using the potential V (X) =

−F0Xe−t2/τ2 which we will treat as a perturbation. Now suppose that, long before the

kick, the oscillator was in its ground state |0〉. Then to first order in the perturbation, the

amplitude for the oscillator to have made a transition into the kth excited state as t→∞is

limt→∞

ak(t) = − i

~

∫ ∞−∞−F0〈k|X|0〉 eikωt−t2/τ2 dt (9.20)

Recalling that X =√

~2mω (A+A†), we see that ak(t) = 0 for k 6= 1, while

limt→∞

a1(t) = F0

√π~

2mωτ e−ω

2τ2/4 (9.21)

where the Gaussian integral is performed by completing the square in the exponent. Thus

the probability of the kick causing a transition is

Prob(|0〉 → |1〉) = F 20

π~2m

τ2

ωe−ω

2τ2/4 (9.22)

– 131 –

Figure 18: As t→∞, the function sin2(xt)/x2t→ πδ(x).

to first order in perturbation theory. This grows like τ2 for small τ , then falls as e−ω2τ2/4

for τ ω−1. For an oscillator of fixed (classical) frequency ω, the maximum probability

occurs when τ ∼ ω−1. Transitions to higher excited states are possible at higher order in

perturbation theory, but if we really kick the oscillator, then perturbation theory will cease

to be appropriate.

9.3 A Constant Perturbation Turned On at t = 0

As a further example, consider the case that

∆(t) =

0 for t < 0

∆ for t > 0 ,(9.23)

where ∆(X,P, · · · ) is a time-independent operator. Let’s start our system so that it’s in

the mth eigenstate |m〉 of H0 at t = 0. Then from (9.19) with t0 = 0 we have

ak(t) = − i

~

∫ t

0〈k|∆|m〉 eiωkmt

′dt′ =

∆km

Ek − Em(1− eiωkmt

), (9.24)

where ∆km = 〈k|∆|m〉 and ωkm = (Ek −Em)/~. Thus the probability of the perturbation

causes the system to be in the kth state at time t is

|ak(t)|2 =4|∆km|2~2ω2

km

sin2(ωkmt/2) . (9.25)

We see that the transition probability depends on the energy difference ~ωkm as well as

the matrix element ∆km.

Consider the graph of (1/Ω2) sin2(Ωt/2) as a function of Ω for fixed t, shown in figure 18.

The height of the central peak is ∝ t2 while its width is ∼ 1/t, where we recall that t is the

time over which the perturbation acted. For large t, |ak(t)|2 is only appreciable for states

|k〉 whose energies obey

t ∼ 2π

|ωkm|=

2π~|Ek − Em|

(9.26)

– 132 –

In other words, if ∆t is the time for which the perturbation has been switched on and ∆E

the energy difference between states which have an appreciable probability of a transition,

then we require

∆t∆E ∼ ~ .For small ∆t, the central peak in figure 18 is broad and we can have appreciable probability

of a transition even between states with ∆E 6= 0. However, if the constant perturbation

has been switched on for a long time, we will only obtain transitions to states |k〉 that are

(very nearly) degenerate with the original states |m〉. In fact as t→∞, for Ω 6= 0 we have

limt→∞

sin2(Ωt/2)

Ω2t≤ lim

t→∞

1

Ω2t= 0 , whilst lim

Ω→0

sin2(Ωt/2)

Ω2t=t

4,

which diverges if we then send t→∞. Furthermore, the integral of this function is∫ ∞−∞

sin2(Ωt/2)

Ω2tdΩ =

π

2,

independent of t. This shows that we can replace

limt→∞

sin2(Ωt/2)

Ω2t=π

2δ(Ω) , (9.27)

so that as t→∞, the transition probability (9.25) behaves as

|ak(t)|2 ∼2π

~2|〈k|∆|m〉|2 δ(ωkm) t . (9.28)

We define the transition rate Γ(m→ k) by

Γ(m→ k) = limt→∞

∂

∂t|ak(t)|2 . (9.29)

In the case of a constant perturbation switched on at t = 0, we thus have

Γ(m→ k) =2π

~|〈k|∆|m〉|2 δ(Ek − Em) (9.30)

where we have written the argument of the δ-function in terms of energies, pulling out a

factor of ~. As claimed, for this constant perturbation we only have a non-zero transition

rate to states which are degenerate with our initial state.

9.4 Fermi’s Golden Rule

Just a small extension of the above gives us one of the most important cases of time-

dependent perturbation theory. Suppose our perturbation takes the form

∆(t) =

0 for t < 0

∆ e−iωt + ∆† e+iωt for t > 0(9.31)

where again ∆ = ∆(X,P, · · · ) is some time–independent operator. Thus, as before our

Schrodinger picture perturbation ∆(t) is turned on at t = 0, but thereafter it now oscillates

– 133 –

with fixed frequency ω. Since ∆(t) appears as a term in the Hamiltonian, it must be

Hermitian, so we’ve included both an e−iωt and e+iωt term and without loss we may take

ω > 0. We’ll see below that these two terms are responsible for different physics. We call

such perturbations monochromatic in anticipation of the case that they describe an applied

electromagnetic field, corresponding to light of frequency ω.

We’ll again consider the case where our system is initially in the mth eigenstate of

the unperturbed H0, so that an(0) = δnm. Performing the time integral in (9.19) in this

monochromatic case is again trivial and yields

ak(t) =〈k|∆|m〉~(ωkm − ω)

[ei(ωkm−ω)t − 1

]+〈k|∆†|m〉~(ωkm + ω)

[ei(ωkm+ω)t − 1

]. (9.32)

This is very similar to our result (9.24), with the time–dependence of the perturbation

just causing ωkm to be replaced by ωkm − ω in the first term and ωkm + ω in the second.

Consequently, as t→∞, we will obtain significant transition amplitudes only to states |m〉for which Ek − Em ≈ ~ω, or else for which Ek − Em ≈ −~ω to high accuracy. The first

situation, where Ek > Em corresponds to our system absorbing energy from the perturba-

tion, whilst the second is stimulated emissions, where the system is prompted to release

energy by the perturbation. In this way, the time–dependent monochromatic perturbation

can be regarded as an enormous source and sink of energy that can be exchanged with the

system.

In the case of absorption, the first term dominates at late times. Thus the probability

that the perturbation excites the system from |m〉 → |k〉 is

|ak(t)|2 ≈4 |〈k|∆|m〉|2~2(ωkm − ω)2

sin2

(ωkm − ω

2t

). (9.33)

with Ek > Em, correct to order |∆|2. Replacing the late–time behaviour of |ak(t)|2 by a

δ-function as before shows that the transition rate is

Γ(m→ k) = limt→∞

∂

∂t|ak(t)|2 =

2π

~|〈k|∆|m〉|2 δ(Ek − Em − ~ω) (9.34a)

in the case that the system absorbs energy from the perturbation. In the opposite case that

the system loses energy back into the perturbation, we have Ek < Em and the corresponding

transition rate is

Γ(m→ k) =2π

~|〈m|∆|k〉|2 δ(Ek − Em + ~ω) . (9.34b)

with an opposite sign in the δ-function. These results were first obtained by Dirac. They

proved to be so useful in describing atomic transitions that Fermi dubbed them ‘golden

rules’. The name stuck and now (9.34a) & (9.34b) are known as Fermi’s golden rule.

In practice, a truly monochromatic perturbation never actually causes transitions be-

tween two discrete energy levels, such as two bound states of an atom, because we cannot

hope to tune its frequency precisely enough so that the δ-functions in (9.34a) or(9.34b) can

be satisfied. There are thus two different classes of circumstances in which Fermi’s golden

rule is useful.

– 134 –

On the one hand, we may be interested not in transitions to some specific final state

|k〉, but to any one of a range of continuum eigenstates of H0. A typical example here

would be transitions between an initial bound state of an atom to any of the continuum of

positive-energy (non-bound) states. In these circumstances, we let n(Ek) δEk denote the

number of (final) states with energy in the interval (Ek, Ek + δEk). The function n(Ek) —

which needs to be calculated in each case — is known as the density of states; note that

it must have dimensions of 1/(energy) in order for n(Ek) δEk to be a number of states.

The total transition probability from |m〉 is then∫|ak(t)|2 n(Ek) dEk, generalizing the sum∑

k |ak(t)|2 in the discrete case. In particular, in the case of absorption the total transition

rate from our initial eigenstate |m〉 is given by∫Γ(m→ k) ρ(Ek) dEk =

2π

~

∫|〈k|∆|m〉|2 δ(Ek − Em − ~ω) ρ(Ek) dEk

=2π

~|〈k|∆|m〉|2 ρ(Ek)

∣∣∣∣Ek=Em+~ω

(9.35)

to lowest non-trivial order in perturbation theory. We see that we always make transitions

through an energy of exactly ~ω, with the density of states telling us how many states of

energy Em + ~ω there actually are.

On the other hand, Fermi’s golden rule is also useful when our perturbation itself

contains a continuum of different frequencies, perhaps thought of as a Laplace or Fourier

transform of some a perturbation with generic time dependence. In this case, transitions

between discrete bound states are possible, because there will always be some perturbation

at exactly the right frequency ω = ωkm = (Ek−Em)/~, even though the rhs takes discrete

values as k varies over different bound states as the final state.

Let’s now consider an example of each of these two cases.

9.4.1 The Photoelectric Effect

First, we’ll use Fermi’s golden rule to calculate the rate at which an applied monochromatic

electromagnetic field can ionize an atom, say hydrogen, stripping off the electron from a

bound state into the continuum of ionized states. We’ll assume that the electromagnetic

field itself may be treated classically – this assumption is valid e.g. in a laser, or at the

focus of the antenna of a radio telescope where the field is large. Of course, if we wish

perturbation theory to produce reliable results, we’ll still need this applied electromagnetic

field to be small compared to the Coulomb fields binding the electron(s) to the nucleus of

the atom.

In the vacuum, the electromagnetic field is divergence free and is entirely generated by

Faraday’s Law ∇× E = −∂B/∂t. The whole electromagnetic field can be described by a

vector potential A via

B = ∇×A and E = −1

c

∂A

∂t. (9.36)

In the monochromatic case, Faraday’s equation is solved by

A(x, t) = ε cos(k′ · x− ωt) , (9.37)

– 135 –

where ε is a constant vector describing the polarization of the wave and k′ is the wavevector.

(We call the wavevector k′, reserving k for later use.) From the vacuum Maxwell equation

∇ ·E = 0, we learn that

k′ · ε = 0 , (9.38)

saying that the wave is transverse. Faraday’s law then reduces to the statement that

ω = c|k′| so that the wave travels at the speed of light.

The coupling of an elementary particle of charge q to such a vector potential is given

by the minimal coupling rule, replacing

P 7→ P− qA(X, t) (9.39)

in the Hamiltonian79, so

H =(P− qA)2

2µ+ V (X) = H0 −

q

2µ(P ·A + A ·P) +

q2A2

2µ, (9.40)

where H0 contains the original potential term V (X) as well as the kinetic term. In our case

of atomic transitions, an electron has charge q = −e so to first–order in e the perturbing

Hamiltonian is

∆(t) =e

2µ

(P · ε cos(k′ ·X− ωt) + cos(k′ ·X− ωt) ε ·P

). (9.41)

The component ε ·P of the momentum operator in the direction of the polarization of the

electromagnetic wave commutes with cos(k′ ·X−ωt) by the transversality condition (9.38)

(recall that ε itself is a constant vector). Thus

∆(t) =e

µcos(k′ ·X− ωt) ε ·P

=e

2µ

(ei(k′·X−ωt) + e−i(k′·X−ωt)

)ε ·P ,

(9.42)

simplifying the perturbation.

Having found the form of our perturbation, we now need calculate its matrix element

between our initial and final states. We’ll consider the case where the hydrogen atom is

initially in its ground state, so that |m〉 = |100〉 with position space wavefunction

ψ0(x) = 〈x|100〉 =1√πa3

0

e−r/a0 ,

where a0 is the Bohr radius. For the final state, we’re interested in the case where the elec-

tron has been liberated from the proton. If the energy of the ejected electron is sufficiently

great, we can ignore its Coulomb attraction back to the ion and treat the final electron as

a free particle of momentum ~k, so we take our final state |k〉 to be the plane-wave

ψk(x) = 〈x|k〉 =1

(2π~)3/2eik·x .

79This is called minimal coupling because more complicated modifications are possible if the charged

particle is not elementary, or in other more complicated circumstances. You’ll learn more about this in the

Part II Classical Dynamics course, or if you take AQM next term.

– 136 –

This transition requires that the electron absorbs energy from the radiation, so the matrix

element we seek comes from the first term on the rhs of (9.42). We have

〈k|∆|100〉 =e

2µ〈k| eik′·X ε ·P |100〉

= − i~e2µ

1

(2π~)3/2(πa30)1/2

∫R3

e−ik·x eik′·x ε ·∇(e−r/a0) d3x .(9.43)

Integrating by parts and using the fact that the bound state wavefunction decays expo-

nentially as |x| → ∞, this is

〈k|∆|100〉 =e~2µ

ε · k(2π~)3/2(πa3

0)1/2

∫R3

e−iq·x e−r/a0 d3x , (9.44)

where q = k − k′ is the momentum transfer and we recall that the polarization vector

is transverse, ε · k′ = 0. We’re left with just the Fourier transform of the ground state

wavefunction. Choosing the z-axis to lie along the direction of q and working in polar

coordinates, this is∫e−iq·x e−r/a0 d3x =

4π

q

∫ ∞0

r e−r/a0 sin(qr) dr =8πa3

0

(1 + a20q

2)2(9.45)

where q = |q|.We now wish to use this matrix element in Fermi’s golden rule for the transition rate.

Since there are a range of possible final momenta, we define the differential rate δΓ|100〉→|k〉to be the rate at which the radiation ionizes an atom so that the momentum of the ejected

electron lies in the range ~(k,k + δk), so

dΓ|100〉→|k〉 =2π

~|〈k|∆|100〉|2 δ(Ek − E100 − ~ω) ~3 d3k

= 2π~2 |〈k|∆|100〉|2 δ(Ep − E100 − ~ω) dφ sin θ dθ k2dk(9.46)

where dΩ = sin θ dθ dφ. To relate this differential rate to our previous expression for the

rate of transition into states of particular energy, recall that we’re treating the final electron

as a free particle, so it’s energy is Ek = ~2k2/2µ. Thus we have

k2dk = k2(dEk/dk)−1dEk = (µk/~2) dEk ,

so for a free particle, the density of states n(Ek) = µk/~2 = µ√

2µEk/~3. Putting all the

pieces together and integrating over the final state energy, the transition rate for electrons

to be emitted into the direction dΩ of solid angle is

dΓ|100〉→|k〉

dΩ=

4e2(ka0)3

πµ~|ε · k|2

(1 + a20q

2)4(9.47)

where ~k =√

2µEk is frozen to√

2µ(~ω + E100) ≈ √2µ~ω and we recall that q = k − k′

is the momentum transfer.

– 137 –

9.4.2 Absorption and Stimulated Emission

Now let’s consider the case where instead of being illuminated by monochromatic light

such as from a laser, our atom is immersed in a bath of thermal radiation. This radiation

bath supplies us with a continuous range of different frequencies, so transitions between

different (discrete) bound states of the atom are now possible.

The problem simplifies if we assume that the typical electromagnetic wavelength 1/k′

in the thermal bath is much larger than the Bohr radius of the atom. This is a good approx-

imation provided the atomic transition occurs between states that are separated in energy

by much less than αµc2, as will be the case for transitions between highly excited energy

levels, perhaps by the valence electrons in a heavy atom. Such wavelengths correspond to

radiation with frequencies that are less than those of soft X-rays. In this approximation,

k′ ·X 1 for all locations in the atom or molecule at which there is significant probability

of finding the electron. To lowest order, we can thus approximate the electric field as being

constant over the scale of the atom. The Hamiltonian for the atom is then

H = H0 + e∑r

E(t) ·Xr , (9.48)

where H0 is the Hamiltonian of the unperturbed atom, E(t) is the fluctuating electric field

due to the radiation and Xr is the position operator for the rth electron in the atom. This

is known as the dipole approximation since the operator e∑

r Xr represents the atom’s

dipole moment. We interpret the transition as occurring due to the interaction between

the applied electric field and this dipole.

Whilst roughly constant over the atom, the electric field E(t) fluctuates rapidly in

time as the atom is jostled by radiation of different frequencies in the thermal bath. In

particular, on average E(t) = 0 since the electric field is equally likely to point in any

direction at any given moment. We take E(t) to be correlated as

Ei(t1) Ej(t2) = δij

∫RP(ω) e−iω(t1−t2) dω . (9.49)

The presence of δij on the rhs means that fluctuations in different directions are uncorre-

lated, whilst fluctuations that are aligned are correlated equally no matter their direction.

This just says that there is no preferred direction for the electric field and holds e.g. for a

thermal bath of radiation.

We can understand the meaning of the function P(ω) as follows. First, note that if

the electric fields are real, then we must have

P(−ω) = P(ω) and P(ω)∗ = P(ω) (9.50)

so that P(ω) is a real, even function of ω. Recall from IB Electromagnetism (or looking

ahead to Part II Electrodynamics) that the energy density of an electromagnetic field is

ε0

2

(E2(x, t) + c2B2(x, t)

)

– 138 –

and in a purely radiative (source–free) field E2 = c2B2. In our case, equation (9.49) shows

that the average energy density we expect to find in the radiation at time t is

ε0Ei(t)Ei(t) = 3ε0

∫ ∞−∞P(ω) dω = 6ε0

∫ ∞0P(ω) dω , (9.51)

so ρ(ω) = 6ε0P(ω) represents the average energy density in the radiation (at any time) due

to frequencies in the range (ω, ω + dω). (If the radiation is truly thermal at temperature

T , then ρ(ω) would be given by Planck’s blackbody formula

ρ(ω) =~ω3

π3c2

1

e−~ω/kBT − 1(9.52)

coming from the Bose–Einstein distribution. You’ll learn about this if you take next term’s

Statistical Physics lectures.)

Now let’s put these results to work. For a transition |m〉 → |k〉, we have from (9.19)

and (9.48) that

ak(t) = − ie

~

∫ t

0

∑r

〈k|E(t′) ·Xr|m〉 eiωkmt′dt′ . (9.53)

We write the probability that the atom is found in state |k〉 at time t as a double integral

|ak(t)|2 =e2

~2

∫ t

0

∫ t

0

∑r,r′

〈k|E(t1) ·Xr′ |m〉 〈m|E(t2) ·Xr|k〉 eiωkm(t1−t2) dt1 dt2 . (9.54)

The correlations (9.49) in our thermal bath shows that, on average,

|ak(t)|2 =e2

~2

∫ t

0

∫ t

0Ei(t1)Ej(t2)

∑r,r′

〈k|(Xr′)i|m〉 〈m(Xr)j |k〉 eiωkm(t1−t2) dt1 dt2

=e2

~2

∣∣∣∣∣∑r

〈k|Xr|m〉∣∣∣∣∣2

×∫ ∞−∞P(ω)

[∫ t

0

∫ t

0ei(ωkm−ω)(t1−t2) dt1 dt2

]dω

=e2

~2

∣∣∣∣∣∑r

〈k|Xr|m〉∣∣∣∣∣2

×∫ ∞−∞P(ω)

∣∣∣∣∫ t

0ei(ωkm−ω)t1 dt1

∣∣∣∣2 dω=

4e2

~2

∣∣∣∣∣∑r

〈k|Xr|m〉∣∣∣∣∣2

×∫ ∞−∞P(ω)

sin2((ωkm − ω)t/2)

(ωkm − ω)2dω ,

(9.55)

where the mod-square includes the Euclidean inner product (dot product) of the two

〈k|Xr|m〉 matrix elements. This form is again familiar, except that instead of a density of

final states, we have P(ω), representing the average energy density ρ(ω) of the frequency ω

component of the radiation. Making our familiar replacement, and performing the integral

over ω using the resulting δ-function, we have the transition rate

Γ(|m〉 → |k〉) =πe2

3ε0~2

∣∣∣∣∣∑r

〈k|Xr|m〉∣∣∣∣∣2

ρ(ωkm) (9.56)

– 139 –

in the case that the states |m〉 and |k〉 are discrete but we have a continuous range of

perturbing frequencies. Again we recall that ωkm = (Ek − Em)/~ is the fixed difference

between the initial and final bound state energy levels.

This result is very intuitive: the rate of transition depends on the probability that the

dipole moment transition can link states |m〉 and |k〉, together with the energy density in

the radiation bath at the right frequency to match this transition. For hydrogenic states of

the atom, we recall from section 6.2.2 that the matrix element vanishes unless the selection

rules

|`− `′| = 1 and |m−m′| ≤ 1

are obeyed. These selection rules were derived under the assumption that the transition is

just due to the dipole approximation where the perturbation involves matrix elements of

the dipole moment operator eX. The effects of both higher orders in perturbation theory,

and inclusion of interactions beyond those of an electric field with wavelength λ a0

mean that transitions that are ‘forbidden’ according to the above selection rules may in

fact occur. They will typically do so, however, at a much smaller rate.

Now, whether or not the radiation bath is thermal, the reality conditions (9.50) show

that the rate Γ(|m〉 → |k〉) in (9.56) is an even function of ωkm. Consequently, for fixed

|Ek − Em|, this rate is the same whether Ek > Em or Ek < Em. For Ek > Em, the atom

absorbs energy from the radiation, exciting to a higher energy level, whilst if Ek < Emthe atom emits energy back into the radiation as it decays to a lower energy state. The

important result we’ve obtained is that the rate for stimulated emission is the same as the

rate of absorption.

9.4.3 Spontaneous Emission

Our treatment above calculated the rate at which an atom would decay (or excite) due to

being immersed in a bath of radiation. However, we currently cannot understand decays

of an isolated atom: if the atom is prepared in any eigenstate of H0, quantum mechanics

says that in the absence of a perturbation it will just stay there. Remarkably, Einstein was

able to relate the rate of spontaneous decay of an atom to the rate of stimulated decay

that we calculated above.

Suppose our atom is immersed in a bath of radiations where the energy density of

photons with frequency in the range (ω, ω + dω) is ρ(ω) dω. Einstein defined Am→k(ωmk)

as the rate at which an atom would spontaneously decay from a state of energy Em to a

state of lower energy Ek, where ωmk = (Em−Ek)/~ is the frequency of the emitted photon.

We let ρ(ωkm)Bk→m(ωkm) denote the rate at which energy is absorbed from the radiation

bath, exciting the atom from |k〉 → |m〉. Similarly, let ρ(ωkm)Bm→k(ωkm) denote the rate

of decay of the excited atom due to stimulation by the presence of the radiation bath.

We calculated Bm→k and Bk→m in the dipole approximation above, finding they were the

same, but let’s pretend that (like Einstein) we don’t know this yet.

In thermodynamic equilibrium these rates must balance, so if there are nk atoms in

state |k〉 and nm in state |m〉 we must have

nm (Am→k(ωmk) + ρ(ωkm)Bm→k(ωkm)) = nk ρ(ωkm)Bk→m(ωkm) . (9.57)

– 140 –

We now borrow two results you’ll derive in the Part II Statistical Mechanics course, next

term. First, when a gas of atoms is in equilibrium at temperature T , the relative numbers

of atoms in states with energies Em and Ek is given by the Boltzmann distribution

nknm

=e−Ek/kBT

e−Em/kBT= e~ωmk/kBT , (9.58)

where kB ≈ 1.38 × 10−23 JK−1 is Boltzmann’s constant. Furthermore, if radiation is in

equilibrium at the same temperature then the density of states at frequency ω is

ρ(ω) =~ω3

π2c3

1

e~ω/kBT − 1, (9.59)

which is essentially the Bose–Einstein distribution. Using these in (9.57) gives

Am→k(ω) =~ω3

π2c3

1

e~ω/kBT − 1

(e~ω/kBT Bk→m(ω)−Bm→k(ω)

). (9.60)

where ω = ωmk. However, Am→k is supposed to be the rate of spontaneous emission of

radiation from the atom, so it cannot depend on any properties of the radiation bath. In

particular, it must be independent of the temperature T . This is possible iff

Bk→m(ω) = Bm→k(ω) (9.61)

so that the rates of stimulated emission and absorption agree. In this case, we have

Am→k(ω) =~ω3

π2c3Bm→k(ω) (9.62)

so that knowing the rate Bk→m(ω) at which a classical electromagnetic wave of frequency

ω = (Em−Ek)/~ is absorbed by an atom, we can also calculate the rate Am→k(ω) at which

the atom spontaneously decays, even in the absence of radiation.

Now, in equation (9.56) we found from a first–principles calculation that

Γ(|m〉 → |k〉) =πe2

3ε0~2

∣∣∣∣∣∑r

〈k|Xr|m〉∣∣∣∣∣2

ρ(ωkm) (9.63)

in the dipole approximation, to first-order in perturbation theory. Consequently, for our

atom Einstein’s B coefficient is

Bk→m(ωkm) = Bm→k(ωmk) =πe2

3ε0~2

∣∣∣∣∣∑r

〈k|Xr|m〉∣∣∣∣∣2

. (9.64)

for stimulated emission or absorption. From Einstein’s statistical argument we thus learn

that

Am→k(ω) =e2ω3

3ε0π~c3

∣∣∣∣∣∑r

〈k|Xr|m〉∣∣∣∣∣2

(9.65)

is the rate at which an excited atom must spontaneously decay, even in the absence of any

applied radiation.

– 141 –

Ingenious though it is, there’s something puzzling about this calculation of the rate

of spontaneous decay. Where does our quantum mechanical calculation – stating that

isolated excited atoms do not decay – go wrong? The answer is that while we treated

the energy levels of the atom fully quantum mechanically, the electromagnetic field itself

was treated classically. Spontaneous emission is only possible once one also quantizes the

electromagnetic field, since it is fluctuations in the zero–point value of this field that allow

the atom to emit a photon and decay; heuristically, the fluctuating value of the quantum

electromagnetic field – even in the vacuum – ‘tickles’ the excited atom prompting the decay.

Treating the EM field classically is appropriate if the radiation comes from a high–intensity

laser, where the energy density of the field is so high that the terms B(ω)ρ(ω) dominate

in (9.57). By contrast, emission of light from the atoms in a humble candle is an inherently

quantum phenomenon, occurring purely by spontaneous emission once the flame is lit —

candlelight shines even in ambient darkness.

Einstein gave his statistical argument in 1916, when Bohr’s model of the atom (the first

thing you met in IB QM) was still the deepest understanding physicists had of quantum

theory. This was at the same time as he was revolutionising our understanding of gravity,

space and time, which obviously wasn’t enough to keep him fully occupied. The quantum

treatment of stimulated emission given in the previous section is due to Dirac in 1926, the

same year as Schrodinger first published his famous equation. A first–principles calculation

of the spontaneous emission rate Am→k, requiring the quantization of the electromagnetic

field, was also given by Dirac just one year later. Things move fast. Dirac’s 1927 results

agreed with those we’ve found via Einstein’s argument. The necessity also to quantize the

electromagnetic field heralded the arrival of Quantum Field Theory.

Let’s use these results to calculate the typical lifetime of an excited state of an isolated

atom. When the radiation density ρ(ω) is very small, the number of atoms nm in the

excited state obeys∂nm∂t

= −Am→k nm (9.66)

so the atom decays exponentially with a characteristic timescale ∼ A−1m→k. For hydrogen or

an alkali atom, typically the outermost electron changes its energy level in the transition,

so we can drop the sum in (9.65) to find

Am→k(ω) =e2ω3

3πε0~c3|〈k|X|m〉|2 (9.67)

where X is the position of the outermost electron. Unless 〈k|X|m〉 vanishes due to some

selection rule, we’d expect this matrix element to be of the typical size a0 = 4πε0~2/µe2

characteristic of the atom, so |〈k|X|m〉|2 ∼ a20. Hence the characteristic lifetime of the

excited atomic state is roughly

τ = A−1m→k ∼

3πε0~c3

e2ω3a20

=3µc3

4~ω3a0(9.68)

where µ is the mass of the electron. Optical light has a frequency of order ω ∼ 1015 Hz,

so the timescale τ is around 107 times longer than the timescale ~/E200 associated to the

– 142 –

energy E200 of the first excited state of hydrogen. Thus around 107 oscillations of the

hydrogen atom occur before it radiates a photon, decaying down into the ground state.

– 143 –